Question

Select all of the ordered pairs that make the inequality y > x2 + 3x – 4 true.

(0, 0)

(2, 1)

(–2, –1)

(–5, –1)

Answer 1

Plug each x into x^2 + 3x − 4 and compare to y:

- (0, 0): RHS = 0^2 + 3·0 − 4 = −4 → 0 > −4 (true)
- (2, 1): RHS = 4 + 6 − 4 = 6 → 1 > 6 (false)
- (−2, −1): RHS = 4 − 6 − 4 = −6 → −1 > −6 (true)
- (−5, −1): RHS = 25 − 15 − 4 = 6 → −1 > 6 (false)

So the pairs that satisfy the inequality are (0, 0) and (−2, −1).